waterfall said:
When you mentioned "field quanta", are you referring to operator field quanta or actual field quanta? This is because as detailed in message 29, the field in QFT are field operator, not the usual field we understood as electromagnetic field for example.
Waterfall, a quantum field is a quantum 'quantity'. In the formalism of quantum physics, these are
operators (or POVMs, which are a related but more complicated object). The 'actual' field IS the
'operator' field.
I'll give you two examples: the total momentum of a system, P, and the electromagnetic field, A.
In CLASSICAL physics, these, or their components in some reference frame, are numbers.
P = {Px, Py, Pz}; A = {phi, Ax, Ay, Az}.
In QUANTUM physics, these are operators. That's a more complicated kind of object. An important
difference with the above case is, operators don't have a value by themselves. This is where the
state comes in in the theory. Quantum states give operators their values (and their indeterminacy).
So, while in classical physics you have A=A(x,y,z,t) as a vector with a definite value assigned
to every point (x,y,z), in quantum physics you have A=A(x,y,z,t) as an operator field, that is,
an operator assigned to every point of space (and time). Once you're given a state you can
assign a value (actually, an expectation value and an indeterminacy) to those operators. If the
indeterminacy is sufficiently small, it can be ignored and you recover the classical field (this
can only happen for fields which do possesses a classical limit, of course. The em field does.)